HW4solutions_3

# HW4solutions_3 - MATH 115 SOLUTION SET 4 ANSWERS TO...

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MATH 115 SOLUTION SET 4 ANSWERS TO SELECTED PROBLEMS § 2.1: 1. a) True, b) False, c) False (look at T : R R deﬁned by T ( x ) = x + 1; but also note that this is not linear.) d) True, e) False (it should be dim V there), f) False ( T could take everything to zero, what a catastrophe!) g) True, h) False, because { x 1 ,x 2 } might not be linearly independent. 10. Since (2 , 3) = - (1 , 0) + 3(1 , 1), we have T (2 , 3) = - T (1 , 0) + 3 T (1 , 1) = - (1 , 4) + 3(2 , 5) = (5 , 11). T is indeed one-to-one. 11. Since (1 , 1) and (2 , 3) are linearly independent, there exists a linear trans- formation T : R 2 R 3 that sends these two vectors to any two vectors in R 3 . Since (8 , 11) = 2(1 , 1) + 3(2 , 3), apply T to both sides to get T (8 , 11) = 2(1 , 0 , 2) + 3(1 , - 1 , 4) = (5 , - 3 , 16). 14. a) Assume T is one-to-one. Let S be a linearly independent subset of V ; we will show T ( S ) is linearly independent. If a 1 T ( v 1 ) + ··· + a n T ( v n ) = 0 with v i V and scalars a i

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## This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.

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HW4solutions_3 - MATH 115 SOLUTION SET 4 ANSWERS TO...

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